Liste des Groupes | Revenir à p relativity |
On Wed, 7 Aug 2024 14:44:15 +0000, Richard Hachel wrote:Yes.>The classical Doppler shift is lambda' = lambda/(1 +/- v). The
Le 07/08/2024 à 16:25, hitlong@yahoo.com (gharnagel) a écrit :>>
Hmm, doesn't look like a laugh. Maybe an OMG! Meaning, you just
realized that Jan is right. Well, maybe a laugh would be appropriate,
too, meaning "how could I have been so wrong!"
>
You come up with your D'=D.sqrt[(1+Vo/c)/(1-Vo/c)], which isn't length
contraction but Doppler shift, which is dependent on the sign of your
Vo. LC is NOT so dependent. It would be a VERY strange universe if
it were.
You say: "it's a Doppler shift".
And for sqrt(1-Vo²/c²)?
Isn't it a Doppler shift?
Yes, it's also a Doppler shift.
relativistic
Doppler equation is lambda' = lambda sqrt(1 - v^2/c^2)/(1 +/- v/c). Which,
of course, is lambda' = lambda sqrt[(1 - v/c)/(1 + v/c) for approaching
and
lambda' = lambda sqrt[(1 + v/c)/(1 - v/c) for receding.
I'm not talking about that.This is what Hachel calls the "internal Doppler effect".It is not a distance effect, and definitely not LC.
Relativists call it the transverse Doppler effect, but the term isNope. The transverse Doppler effect is simply time dilation. Your
neither fair nor pretty.
equation has longitudinal Doppler built into it.
Mais bordel, vous allez être poli, oui? Je vous dis, bordel de merde, que :The longitudinal Doppler effect is already a relativistic effect.Nope.
MERDE !!!When Römer observes the moons of Jupiter, his measurements are correct:Nope. There are only longitudinal and transverse.
but he will say: "When you cut a dog's legs, it no longer comes when you
hit its bowl to eat: cutting a dog's legs affects its eardrums".
I would prefer that we speak of internal Doppler effect, and external
Doppler effect. The terms would be more accurate.
Les messages affichés proviennent d'usenet.